Which adaptation might protect a plant from a predator species?

Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission lin

Lerok [7]

Answer:

A. z (galaxy 1) = -0.00515, z (galaxy 2) = -0.01707, z (galaxy 3) = -0.04008

B. v (galaxy 1) = -0.00517c, v (galaxy 2) = -0.01707c, v (galaxy 3) = -0.0401c

C. d (galaxy 1) = -21 Mpc,d (galaxy 2) = -69.4 Mpc, d (galaxy 3) = -163 Mpc

Explanation:

We begin by listing out the parameters we were given:

λ (obsv) = 656.3 nm, λ (emit 1) = 659.7 nm, λ (emit 2) = 667.7 nm,

λ (emit 3) = 683.7 nm

A) Using the Redshift formula, we have:

z = [λ (obsv) - λ (emit)] ÷ λ (emit)

For galaxy 1:

z = [λ (obsv) - λ (emit 1)] ÷ λ (emit 1)

z = (656.3 - 659.7) ÷ 659.7 = -0.00515

z = -0.00515

For galaxy 2:

z = [λ (obsv) - λ (emit 2)] ÷ λ (emit 2)

z = (656.3 - 667.7) ÷ 667.7 = -0.01707

z = -0.01707

For galaxy 3:

z = [λ (obsv) - λ (emit 3)] ÷ λ (emit 3)

z = (656.3 - 683.7) ÷ 683.7 = -0.04008

z = -0.04008

B) Using the Doppler formula, we have:

(Δλ ÷ λ) = v ÷ c

v = c * (Δλ ÷ λ)

but, z = (Δλ ÷ λ)

⇒ v = c * z

speed of light (c) = 3 x $10^{8}$ m/s

For galaxy 1:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.00515)

v = -1.55 x $10^{6}$ m/s

v = -0.00517c

For galaxy 2:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.01707)

v = -5.12 x $10^{6}$ m/s

v = -0.01707c

For galaxy 3:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.04008)

v = -12.03 x $10^{6}$ m/s

v = -0.0401c

N.B: the negative value of velocity connotes that the galaxies are moving away from us (not towards us)

C) Using Hubbleʹs law, we have:

v = H · d

where:

v = velocity of a galaxy (km/s), d = distance (Mpc),

H = Hubble's constant (km/s/Mpc) = 73.8 km/sec

d = v ÷ H

We use the velocities calculated in B) above

For galaxy 1:

v = -1.55 x $10^{3}$ km/s

d = -1.55 x $10^{3}$ ÷ 73.8

d = -21 Mpc

For galaxy 2:

v = -5.12 x $10^{3}$ km/s

d = -5.12 x $10^{3}$ ÷ 73.8

d = -69.4 Mpc

For galaxy 3:

v = -12.03 x $10^{3}$ km/s

d = -12.03 x $10^{3}$ ÷ 73.8

d = -163 Mpc

N.B: distance cannot be a negative value

4 0

3 years ago

Which of the following statements would best describe a biome?

muminat

Answer: D

Explanation:

5 0

2 years ago

It's urgent please help! find the perimeter of the triangle with vertices listed (-5,2), (3,2), (1,6)

vovangra [49]

Answer:

19.68

Explanation:

Given:

Vertices : (-5,2), (3,2), (1,6)

To get the length of each side, we need to

Length = √(x2 - x1)^2 + (y2 - y1)^2

calculate the distance between the points ;

Taking, (-5,2), (3,2)

L1 = √(3 -(-5))^2+ (2 -2)^2

L1 =√(3 + 5)^2 + 0^2

L1 = √8^2 = √64 = 8

Taking L2 ; (3,2), (1,6)

L2 = √(1 -3)^2 + (6 - 2)^2

L2 = √ (-2)^2 + 4^2

L2 = √ 4 + 16

L2 = √20 = 4.47

Taking L3 ; (-5,2), (1,6)

L3 = √(1 - (-5))^2 + (6 - 2)^2

L3 = √(1 + 5)^2 + (6 - 2)^2

L3 = √6^2 + 4^2

L3 = √36 + 16 = √52

L3 = 7.21

Therefore, perimeter = L1 + L2 + L3

8 + 4.47 + 7.21 = 19.68

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